## Precalculus (6th Edition)

$\sqrt 3$
Sum of Internal angles of triangle is $180^0$ Then $\angle C=180-\left( 60+75\right) =45$ From the laws of sines $\dfrac {\sqrt {2}}{\sin \angle C}=\dfrac {a}{\sin 60}\Rightarrow a=\dfrac {\sqrt {2}\times \sin 60}{\sin 45}=\dfrac {\sqrt {2}\times \dfrac {\sqrt {3}}{2}}{\dfrac {\sqrt {2}}{2}}=\sqrt {3}$